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anonymous

  • 5 years ago

The second derivative of (4-x^2)*exp(-1.16x^2)

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  1. anonymous
    • 5 years ago
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    can anyone help me with this?? =]

  2. anonymous
    • 5 years ago
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    you can use the product rule right?

  3. anonymous
    • 5 years ago
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    h(x)=f(x)g(x) h(x)=f(x)g(x)=f(x)g(x)+f(x)g(x)

  4. anonymous
    • 5 years ago
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    h'(x) = [f(x)g(x)]' = f'(x)g(x)+f(x)g'(x)

  5. anonymous
    • 5 years ago
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    yes but it doesn't work. my online program does not accept it. (4-x^2)(5.3824*^2*exp(-1.16x^2)-(-2.32xexp(-1.16x^2)(-2x)-(exp(-1.16x^2)(-2)-(-2x)(-2.32xexp(-1.16x^2)

  6. anonymous
    • 5 years ago
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    is that what you got?

  7. anonymous
    • 5 years ago
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    I didn't work it out. I was just pointing out that you can use the product rule for this kind of problem.

  8. anonymous
    • 5 years ago
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    i did. do you think you can check my work?

  9. anonymous
    • 5 years ago
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    but your solution seems correct. see if you can simplify it further. Maybe your online tool will accept it then

  10. anonymous
    • 5 years ago
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    I think I am getting a slightly different answer from you. check your calculations.

  11. anonymous
    • 5 years ago
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    what is slightly different about it? I just double checked!

  12. anonymous
    • 5 years ago
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    did you simplify it. wait let me post my solution

  13. anonymous
    • 5 years ago
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    you can set x^2 = y and factor the second term.

  14. anonymous
    • 5 years ago
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    as a quadratic equation of course.

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